Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $z = \dfrac{-4n + 8}{n - 5} \div \dfrac{n^2 - 11n + 18}{-6n + 30} $
Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{-4n + 8}{n - 5} \times \dfrac{-6n + 30}{n^2 - 11n + 18} $ First factor the quadratic. $z = \dfrac{-4n + 8}{n - 5} \times \dfrac{-6n + 30}{(n - 2)(n - 9)} $ Then factor out any other terms. $z = \dfrac{-4(n - 2)}{n - 5} \times \dfrac{-6(n - 5)}{(n - 2)(n - 9)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ -4(n - 2) \times -6(n - 5) } { (n - 5) \times (n - 2)(n - 9) } $ $z = \dfrac{ 24(n - 2)(n - 5)}{ (n - 5)(n - 2)(n - 9)} $ Notice that $(n - 5)$ and $(n - 2)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ 24\cancel{(n - 2)}(n - 5)}{ (n - 5)\cancel{(n - 2)}(n - 9)} $ We are dividing by $n - 2$ , so $n - 2 \neq 0$ Therefore, $n \neq 2$ $z = \dfrac{ 24\cancel{(n - 2)}\cancel{(n - 5)}}{ \cancel{(n - 5)}\cancel{(n - 2)}(n - 9)} $ We are dividing by $n - 5$ , so $n - 5 \neq 0$ Therefore, $n \neq 5$ $z = \dfrac{24}{n - 9} ; \space n \neq 2 ; \space n \neq 5 $